5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book May 8, 2018 13:46

Review of Precalculus 81

coordinates ofMto be (2, 3). The slope

of medianAMis

3

4

. Thus, an equation

ofAMisy− 3 =

(

3

4

)

(x−2).

3. Since the circle is tangent to the line
   y=−1, the radius of the circle is 2 units.
   Therefore, the equation of the circle is
   (x−2)^2 +(y+3)^2 =4.


4. The two derived equations are
   x− 2 = 2 x+5 andx− 2 =− 2 x−5. From
   x− 2 = 2 x+5,x=−7 and from
   x− 2 =− 2 x−5,x=−1. However,
   substitutingx=−7 into the original
   equation|x− 2 |= 2 x+5 results in
   9 =−9, which is not possible. Thus, the
   only solution is−1.


5. The inequality| 6 − 3 x|<18 is
   equivalent to− 18 < 6 − 3 x<18. Thus,
   − 24 <− 3 x<12. Dividing through by
   −3 and reversing the inequality sign, you
   have 8>x>−4or− 4 <x<8.


6. Since f(x+h)=(x+h)^2 +3(x+h), the

expression

f(x+h)−f(x)

h

is equivalent

to

[(x+h)^2 +3(x+h)]−[x^2 + 3 x]

h

=

(x^2 + 2 xh+h^2 + 3 x+ 3 h)−x^2 − 3 x

h

=

2 xh+h^2 + 3 h

h

= 2 x+h+ 3

7. The graph of equation (2) 4x^2 + 9 y^2 = 36
   is an ellipse, and the graph of equation
   (4)y^2 −x^2 =4 is a hyperbola intersecting
   they-axis at two distinct points. Both of
   these graphs fail the vertical line test.
   Only the graphs of equations (1)xy=− 8
   and (3) 3x^2 −y=1 (which are a
   hyperbola in the second and fourth
   quadrants and a parabola, respectively)
   pass the vertical line test. Thus, only
   (1)xy=−8 and (3) 3x^2 −y=1 are
   functions.


8. The domain ofg(x)is− 5 ≤x≤5, and
   the domain off(x) is the set of all real

numbers. Therefore, the domain of

(f ◦g)(x)=

(√

25 −x^2

) 2

= 25 −x^2 is

the interval− 5 ≤x≤5.

9. From the graph,
   (f −g)(2)=f(2)−g(2)= 1 − 1 =0,
   (f ◦g)(1)=f(g(1))=f(0)=1, and
   (g◦f)(0)=g(f(0))=g(1)=0.


10. Lety= f(x) and, thus,y=x^3 +1.
    Switchxandyto obtainx=y^3 +1.
    Solve fory, and you will have
    y=(x−1)^1 /^3. Thus,f−^1 (x)=(x−1)^1 /^3.


11. The amplitude is 3, frequency is^1 / 2 , and
    period is 4π. (See Figure 5.8-1.)

− 3

− 1

− 2

0

y = 3 cos ( )

x

y

− 2 π −1. 5π − 1 π−0.5π 0.5π 1 π 1. 5π 2 π

1

2

3

1

2 x

Figure 5.8-1

12. Note that (1)y=lnxis the graph of the
    natural logarithmic function. (2)y =
    ln(−x) is the reflection about they-axis.
    (3)y=−ln(x+3) is a horizontal shift
    2 units to the left followed by a reflection
    about thex-axis. (See Figure 5.8-2.)

− 6

− 9

− 3

− 4 − 2 0 642

y = ln x

y = −ln(x + 2)

y = ln(−x)

x

y

− 6

9

3

6

Figure 5.8-2